=lim(n tends to infinity) (cosx/2 * cosx/4 * ... * cosx/2n * sinx/2n)/sin (x/2n)

Obviously, judging from the double angle formula,

2sinacosa=sin2a

therefore

Cosx/2 n * sinx/2 n = [Sinks/2 (n-1]]/2

cosx /2^(n- 1)* sinx /2^(n- 1)=[sinx /2^(n-2)]/2

And so on,

(cosx/2 * cosx/4 *…*cosx/2^n *sinx/2^n)

=(sinx) / 2^n

therefore

Primitive restriction

=lim(n tends to infinity) [(sinx)/2 n]/sin (x/2 n)

=lim(n tends to infinity) (sinx)/[2 n * sin (x/2 n)]

=lim(n tends to infinity) [(sinx)/x]/[sin (x/2 n)/(x/2 n)]

Obviously, 2 n tends to infinity, that is, x/2 n tends to 0,

Then sin (x/2 n)/(x/2 n) tends to 1,

that is

Original limit = sinx /x